# Simple circuit to measure capacitance.

Joined Nov 12, 2018
8
Hi. I am currently trying to use a capacitance sensor to measure percentage of air in a air-water mixture. The sensor itself is just two aluminum foils attached outside a pvc pipe. The outer side of the aluminum foils are shielded so that only electrical field disturbances in the pipe is measured.

I have seen and tried many circuits, all of them quite complicated for me (I am a mechanical engineering graduate with very limited circuit knowledge) and I am wondering why a simple circuit (Figure 1) would not be used?

Value of R1 is set based on the range of capacitance that needs to be measured from C1. So, fluctuating value from capacitance sensor C1 will induce different voltage drop for R1, and the diode and C2 straightens the signal so that the final measured voltage is a rectified signal.

So my question is, is there any downside of using this circuit? I know it works because I used it, but I can't help to wonder if people with good electronics and circuit background know why this simple circuit is not used by others.

Figure 1

#### Sensacell

Joined Jun 19, 2012
2,683
Here is a great trick.

Convert the change in capacitance to a change in frequency, then use digital techniques to measure the frequency.
A $2.00 micro-controller with a quartz crystal oscillator can easily measure part-per-million changes in frequency, by comparing the incoming signal to the crystal oscillator. Looking for minute changes in capacitance, the resolution you will need makes analog techniques very difficult. A simple RC oscillator circuit, with your sensor as the "C" can create the signal, just make sure the power supply is clean and very stable. #### BR-549 Joined Sep 22, 2013 4,938 "to measure percentage of air in a air-water mixture." What does that mean? What are you measuring? "The sensor itself is just two aluminum foils attached outside a pvc pipe. The outer side of the aluminum foils are shielded so that only electrical field disturbances in the pipe is measured." What does that mean? Better way to do what? Humidity? Diffuser liquor? Dissolved air? Rain? What does a ME call a air water mixture? Last edited: Thread Starter #### azimarshad Joined Nov 12, 2018 8 "to measure percentage of air in a air-water mixture." What does that mean? What are you measuring? "The sensor itself is just two aluminum foils attached outside a pvc pipe. The outer side of the aluminum foils are shielded so that only electrical field disturbances in the pipe is measured." What does that mean? Better way to do what? Humidity? Diffuser liquor? Dissolved air? Rain? What does a ME call a air water mixture? This is what I was trying to explain: To be more precise, I am measuring the void fraction of the air-water mixture in the pipe. Depending on the electrode shape, I can measure volumetric void fraction or local void fraction that is transient. My question was more about the circuit I was using. Although I know that the circuit works, I wonder if there are any disadvantages of using the circuit since I have not seen anyone else use it. Thread Starter #### azimarshad Joined Nov 12, 2018 8 Here is a great trick. Convert the change in capacitance to a change in frequency, then use digital techniques to measure the frequency. A$2.00 micro-controller with a quartz crystal oscillator can easily measure part-per-million changes in frequency, by comparing the incoming signal to the crystal oscillator.

Looking for minute changes in capacitance, the resolution you will need makes analog techniques very difficult.
A simple RC oscillator circuit, with your sensor as the "C" can create the signal, just make sure the power supply is clean and very stable.
Thank you so much for the quick reply. Will this work if the capacitance changes with time? I mean if I need a profile of the capacitance changes over time, will I be able to measure it?

#### KeepItSimpleStupid

Joined Mar 4, 2014
4,377
What's the order of magnitude of Vr and the AC voltage?

It is nothing mor etha an RC filter made from R and the sensor C.
The half wave rectified and filtered again.

The point of the circuit is to illustrate what needs to happen with IDEAL components.

The diode has a diode drop and really doesn't act as a diode at very low currents. It acts more like a resister. So, possibly a "precision rectifier" is needed there. if the diode is silicon, you have a temperature dependent voltage drop around 0-6-0.7V at room temperature to deal with.

Real Capacitors do have an inherent series resistance and a parallel resistance across them.

Your voltmeter likely has a 10 Meg Ohm input Z which influences the circuit too. So, it looks like a 10 M resistor. Yes, you can get better meters.

The question is, how close to ideal do the components have to be? If the excitation is 120 V, then 0.6 V drop doesn't matter much. If it were 1VAC, it would.

The type of capacitor and the tolerances might matter too.

How do you calibrate the thing? Oscillator level?

#### Sensacell

Joined Jun 19, 2012
2,683
The circuit shown is mostly conceptual, it's not the way to do it in the real world.
Nobody would use it because it presents many difficulties that are not easy to overcome.

1) The oscillator needs have extreme amplitude stability, any variation here would swamp your signal.
This is not easy to accomplish.

2) The static zero output point would drift all over the place. you would need a way to fix this.

3) the single diode detector has unpleasant non-linearity that messes things up still further.
That's what post #6 refers to.

For the time-based measurements, you can have a short sample period, ( a few milliseconds ) but the longer you sample, the more precise the measurement can be.

Beware of "simple" looking circuits, they are traps for beginners.
More complicated circuits are usually based on principles that inherently compensate for subtle but really annoying problems.

#### BR-549

Joined Sep 22, 2013
4,938
That interesting. It works and you have used it. What kind of voltage do you get at 100% water and what do you get with 100% air?

Is the slope linear between the two?

#### MrChips

Joined Oct 2, 2009
22,526
There are many other ways of measuring capacitance. Many circuits are dependent on capacitance, such as a 555-timer, MC14538 monostable multivibrator, an oscillator circuit, or a simple R-C on a GPIO pin of a MCU.

Joined Nov 12, 2018
8
What's the order of magnitude of Vr and the AC voltage?

It is nothing mor etha an RC filter made from R and the sensor C.
The half wave rectified and filtered again.

The point of the circuit is to illustrate what needs to happen with IDEAL components.

The diode has a diode drop and really doesn't act as a diode at very low currents. It acts more like a resister. So, possibly a "precision rectifier" is needed there. if the diode is silicon, you have a temperature dependent voltage drop around 0-6-0.7V at room temperature to deal with.

Real Capacitors do have an inherent series resistance and a parallel resistance across them.

Your voltmeter likely has a 10 Meg Ohm input Z which influences the circuit too. So, it looks like a 10 M resistor. Yes, you can get better meters.

The question is, how close to ideal do the components have to be? If the excitation is 120 V, then 0.6 V drop doesn't matter much. If it were 1VAC, it would.

The type of capacitor and the tolerances might matter too.

How do you calibrate the thing? Oscillator level?

So the AC voltage is 10V, and VR ranges from 0.008V to 0.314V (and these values fluctuates with time). Now I understand that the circuit I used is an ideal circuit and it does not take into consideration voltage drop etc. And heavy voltage fluctuation was one of the problems that I had when doing experiment using this circuit.

For calibration, I don't really need to know what the capacitance values are. I am more interested in what voltage values correspond to what void fraction in the pipe. So what I hope to achieve is to have a calibration curve between void fraction and voltage, and use is for measurement in many other pipes.

Joined Nov 12, 2018
8
The circuit shown is mostly conceptual, it's not the way to do it in the real world.
Nobody would use it because it presents many difficulties that are not easy to overcome.

1) The oscillator needs have extreme amplitude stability, any variation here would swamp your signal.
This is not easy to accomplish.

2) The static zero output point would drift all over the place. you would need a way to fix this.

3) the single diode detector has unpleasant non-linearity that messes things up still further.
That's what post #6 refers to.

For the time-based measurements, you can have a short sample period, ( a few milliseconds ) but the longer you sample, the more precise the measurement can be.

Beware of "simple" looking circuits, they are traps for beginners.
More complicated circuits are usually based on principles that inherently compensate for subtle but really annoying problems.

Now I understand why the circuit is not used. I have done some measurements and although it worked, I have all the problems that you listed above but I do not know what caused them. What do you think of the capacitance bridge in post #10? Is that still a conceptualized circuit?

Joined Nov 12, 2018
8
That interesting. It works and you have used it. What kind of voltage do you get at 100% water and what do you get with 100% air?

Is the slope linear between the two?
The values of VR at 100% water and 100% air are 0.314V and 0.008V, and these time averaged values change with time and the increase or decrease is not linear. Although I managed to run some experiments, it is not efficient and not reliable in the long run since I have to calibrate the water and air values before every single measurement.

This is the calibration curve that I get, and it seems reasonable with the type of circuit that I used.

Here, normalized voltage is (VR-Vair)/(Vwater-Vair)

Joined Nov 12, 2018
8
C bridge approach stim-ed with AC to prevent electrode deposition (if electrodes
in contact with fluid).

Do you know range of C sensor generates ?

Regards, Dana.

From the calculation and measurements that I made, I would say that the C sensor ranges between 0.05 pF to 0.001 pF.

For my case, the electrodes are not in contact with water, so there is no electrode deposition problem.

Looking at the explanation of the capacitance bridge circuit, I assume that it is not suitable for a transient measurement since one of the resistor needs to be adjusted to obtain a balanced bridge?

#### MrChips

Joined Oct 2, 2009
22,526

From the calculation and measurements that I made, I would say that the C sensor ranges between 0.05 pF to 0.001 pF.

For my case, the electrodes are not in contact with water, so there is no electrode deposition problem.

Looking at the explanation of the capacitance bridge circuit, I assume that it is not suitable for a transient measurement since one of the resistor needs to be adjusted to obtain a balanced bridge?
Capacitances between 0.05pF and 0.001pF would be next to impossible to measure.
You cannot normally measure less than 1pF.

Joined Nov 12, 2018
8
Capacitances between 0.05pF and 0.001pF would be next to impossible to measure.
You cannot normally measure less than 1pF.
I see. Must be calculation error in my part because I based the calculation on a simple RC circuit connected in series, with the voltage reading from my measurements. I used a 100k resistor and 10V AC, and my resistor voltage reading was in the range of 0.008V to 0.314V. I back calculated the Imax and then Z, and then C from there.

#### Lemi çağlıbektaş

Joined Feb 8, 2018
2
FDC2214 from texas instruments.

#### bassbindevil

Joined Jan 23, 2014
147
There's the circuit that AADE came up with, originally with logic chips, then a PIC. From http://www.morsex.com/aade/lcmeter.htm, it "works by measuring the shift in frequency caused by inserting an unknown into its oscillator tank circuit" Frequency is a whole lot easier to measure accurately (with digital circuits) than amplitude. More info/links here: https://wa2mze.wordpress.com/2018/11/12/a-digital-l-c-meter/
I don't know if the time required to measure frequency would be a problem or not.

#### wayneh

Joined Sep 9, 2010
16,520
There's the circuit that AADE came up with, originally with logic chips, then a PIC. From http://www.morsex.com/aade/lcmeter.htm, it "works by measuring the shift in frequency caused by inserting an unknown into its oscillator tank circuit" Frequency is a whole lot easier to measure accurately (with digital circuits) than amplitude. More info/links here: https://wa2mze.wordpress.com/2018/11/12/a-digital-l-c-meter/
I don't know if the time required to measure frequency would be a problem or not.
Old thread resurrected by a new member. Caught me too.

#### Janis59

Joined Aug 21, 2017
1,245
Strange but I bought such a device, nice, accurate, fabrique made in Biltema some years ago, designed to measure the wood dryness. Just push the two needles into wood plak few mm deep and reading is ready with sensitivity better than 0,05%. As it price was sth between 2 and 3 Eur I wonder is there any rational reason to make such playmate by own???